The question goes like this:
Suppose is a bounded function such that $f$ is Riemann integrable on $[a,1]$ for every $a \in (0,1)$. Is $f$ Riemann integrale on $[0,1]$?
My approach:
Let us split the domain of $f$, i.e $[0,1]$ into two main 'almost disjoint' subintervals $[0, \xi]$ and $[\xi, 1]$. Evidently, $ \forall \xi \in (0,1)$, $f: [\xi,1] \rightarrow \mathbb{R}$ is integrable. Therefore, by the necessary and sufficient condition for integrability for $f$ on $[\xi,1]$ we can say, for each $ \epsilon/2 >0$, $\exists$ a partition $P^*$ of $[\xi,1]$such that $0 \leq U(P^*,f)-L(P^*,f) < \epsilon/2$ . Now consider $\mathbf{ sup_{[0, \xi]} \ f}=M_\xi$ and $\mathbf{ inf_{[0, \xi]} \ f}=m_\xi$. Now consider the main partition $P$ of $[0,1]$, i.e $\{\xi\} \cup P^*$. The upper sum minus the lower sum of $f$ over its domain $0 \leq U(P^*,f)-L(P^*,f) + (M_\xi - m_\xi)\xi < \epsilon/2+(M_\xi - m_\xi)\xi$ . Choosing $\xi= \frac{\epsilon}{2 (M_\xi - m_\xi)}$, we get $0 \leq U(P,f)-L(P,f) < \epsilon$, which implies the integrability of $f$ on $[0,1]$.
Kindly verify the validity of the proof.